Dynamics - forces in two objects, Newton's third law

In summary: What "vectors" are you talking about? Can you give me an example?In summary, the conversation discusses a problem involving two buckets of nails being pulled up to a roof by a rope. Each bucket has a mass of 5.0kg and the action-reaction force between them is 60N. The conversation includes a free body diagram and equations for calculating the acceleration and force applied by the worker lifting the buckets. However, there is confusion about the terminology and the answer in the back of the textbook stating the force applied as 71N [N] does not match with the calculations. The person summarizing suggests that the author of the answer may have
  • #1
shaools
18
0

Homework Statement



Two buckets of nails are hung one above the other and pulled to a roof by a rope. Each bucket of nails has a mass of 5.0kg. The action-reaction force between the buckets is 60N. Calculate the acceleration and the force applied by the worker lifting them up.

Given: (draw a fbd)
mA= 5kg
FB on A= 60 N

mB= 5kg
FA on B= 60 N [N]

ag = 9.8m/s2

Required:
acceleration(a) and force applied(Fapp)

Homework Equations


Fnet=ma
Fnet= Fapp+(-Fg)

The Attempt at a Solution



So, I need to calculate the net forces of one of the pails in order to find acceleration, because the pails have the same acceleration whether or not you consider them as separate objects or as apart of one.

Working with B:
Fnet= Fapp+(-Fg)
Fnet= FA on B+ (mB*ag)
Fnet= 60N + (5kg*-9.8m/s2)
Fnet= 60N + 49N
Fnet= 11N

a= Fnet/mB
a= 11N/5kg
a= 2.2m/s2

Now, I'm not sure how to approach calculating the applied force.
Does the force acting down on A include both the reaction force of 60N, as well as the force of gravity?
I tried plugging in these values into the equation F=ma, so F=10kg*2.2m/s2=22N, but I think this value is the net force of both of the pails together and not just the applied force? The applied force should be much larger but I'm not sure how to find it.

The answer at the back of the textbook is 71N [N].
Help would be very appreciated, my classmates and I were stumped with this all class !
 
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  • #2
This one stumps me, as well. Are you sure you found the correct answer in the back of the book? What is the answer for the acceleration.

My assumptions:
- A vertical rope with tension Fapp pulls up on a bucket A.
- A vertical rope with tension T2 (60 N) pulls down on bucket A and up on bucket B.
- Gravity pulls down on bucket A with force mAg (5 kg x 9.8 m/s^2) and down on bucket B with force mBg (5 kg x 9.8 m/s^2).
- Both buckets A and B accelerate upwards with the same acceleration a.
- The question asks for Fapp and a.

I do not get Fapp = 71 N.
 
  • #3
Hm, I have a few more questions.

Should the force of gravity be doubled at bucket B because its got the mass of both buckets to pull down at that point?

And also, when Fapp is applied to bucket A, how is that force translated to bucket B? This goes for any question with a force being applied to one of two connected objects, which make up most of my questions for tonight. I get confused on the mathematics when shifting the applied for to the second object.
 
  • #4
Oh and yes I'm sure about the answer being 71N, since its listed with the acceleration which is 2.2m/s^2, which is correct.
 
  • #5
shaools said:
Should the force of gravity be doubled at bucket B because its got the mass of both buckets to pull down at that point?
No. The mass of bucket B is the mass at bucket B. Gravity acts on the buckets individually and independently. Thus, you need a fbd for each bucket.

shaools said:
And also, when Fapp is applied to bucket A, how is that force translated to bucket B?
Well, that's what is confusing me about this problem statement. I have never seen a problem like this stated like this.

I am assuming that the "action-reaction force" is the tension in the rope that connects A and B, i.e. 60 N pulls up on bucket B, and so, by Newton's third law, bucket A experiences a downward pull of 60 N (in addition to A's own weight) from the rope that is connected to B. So, effectively, the person pulling up A feels the weight of A and the tension in the connecting rope combined.
 
  • #6
shaools said:
... the acceleration ... is 2.2m/s^2, which is correct.
Yes. OK. That is the answer that I got for the acceleration. This may be just an issue with terminology, then. I will try to figure out what the terminology means, and see if any force at all in the problem turns out to be 71 N. Are you sure that you're not neglecting to mention anything ...
 
  • #7
Ah okay, thank you.

Say I have another question like this but I'm working with only one object, and all the forces I'm working with are vertical, how does acceleration work in with each of the vectors?
Like for example, I have an object weighing 5kg, and gravity is pulling down on it at 9.8m/s^2, the downward force would be 49 N right? Now, say it was being pulled upwards at an acceleration of 10m/s^2, do I work gravity into my calculations when I'm trying to determine the force acting upwards on the object? Or do I just use 10m/s^2?

Geez, I'm just really confused by this whole concept.
 
  • #8
turin said:
Are you sure that you're not neglecting to mention anything ...

Lol, well the answer exactly is 71 N [N].

But aside from that, no, I'm sure this is all the information I've been given.

..Oh, but, when I tried asking my teacher for help, he seemed to think that the 60 N that was noted in the question was the net force, even though it is stated as the action-reaction force. So, I'm not sure if he made a mistake or if I just have a warped idea of what "net force" is.
 
  • #9
For the bottom bucket 60-5*9.8=5a from which a=2.2m/s^2
For the top bucket F-60=5*2.2 from which F=71N
 
  • #10
The only thing that I can guess is that, for some reason, the author of the answer added the "action-reaction force" (60 N) to the resultant force on bucket A (ma = 11 N). I have no idea why the author would do that.

shaools said:
Say I have another question like this but I'm working with only one object, and all the forces I'm working with are vertical, how does acceleration work in with each of the vectors?
I don't know what you mean. The acceleration is a resultant quantity that takes into account all of the force vectors. Simply add all of the applied forces in order to obtain the resultant force. Then, divide by the mass of the object to obtain the acceleration. This can be straightforward (e.g. in celestial mechanics) or it can be confounded by contact/constraint forces that do not exist without a reaction (e.g. sliding blocks on surfaces with friction).

shaools said:
... for example, I have an object weighing 5kg, and gravity is pulling down on it at 9.8m/s^2, the downward force would be 49 N right?
Yep.

shaools said:
Now, say it was being pulled upwards at an acceleration of 10m/s^2, do I work gravity into my calculations when I'm trying to determine the force acting upwards on the object?
Absolutely. If you assume that gravity is relevant to the problem, then you must incorporate it into Newton's second law. The resultant (or net) force that appears in Newton's second law is the vector addition of every force acting on the body. However, as I mentioned, sometimes this can be less than straightforward.
 
  • #11
Dadface said:
For the bottom bucket 60-5*9.8=5a from which a=2.2m/s^2
For the top bucket F-60=5*2.2 from which F=71N
I assumed that there are two downward forces acting on the top bucket: "action-reaction" and weight. What happened to the weight of the top bucket?
 
  • #12
As I said to turin privately, the acceleration of 2.2 m/s^2 is consistent with a different answer for the total applied force (NOT 71 N). So, I agree with him.
 
  • #13
turin said:
I assumed that there are two downward forces acting on the top bucket: "action-reaction" and weight. What happened to the weight of the top bucket?

Agreed.

EDIT: not to mention the fact that: [ 71 N - (m1 + m2)g ] / (m1 + m2)

is NOT equal to 2.2 m/s^2. Heck, the net force is not even upward in that case!

EDIT 2: In fact I think that settles it. The worker must be applying an upward force greater than (m1 + m2)g if the buckets are moving upward. Turin and I are right.
 
  • #14
Thanks a bunch turin. I'm really fuzzy with this stuff so I keep coming up with more questions than answers :/.

And yeah, trying to work backwards from 71N only gave me answers where I had to ignore some component or another of the question.

shaools said:
Say I have another question like this but I'm working with only one object, and all the forces I'm working with are vertical, how does acceleration work in with each of the vectors?

O messy question. I guess I just meant a question where I'm given an object and an acceleration, but have an opposing force like friction or gravity present. How does that friction or gravity translate into the mathematics for the acceleration, or the applied force, or whatever.
 
  • #15
turin said:
I assumed that there are two downward forces acting on the top bucket: "action-reaction" and weight. What happened to the weight of the top bucket?

Hello turin,I understood the problem as 60N being the total force pulling down on the top bucket assuming that this includes the weight.However it does seem odd somehow and I need to clarify my thoughts on it
 
  • #16
Dadface said:
Hello turin,I understood the problem as 60N being the total force pulling down on the top bucket assuming that this includes the weight.However it does seem odd somehow and I need to clarify my thoughts on it

Then aren't you interpreting the 60 N inconsistently? For the lower bucket, you assume it is the tension in the rope, and doing so gives you a value for a.

For the upper bucket, you suddenly assume that the 60 N is the tension in the rope PLUS the weight of the bucket? No wonder the answer for a is not consistent with the answer for Fapp. See my previous post.
 
  • #17
If you guys aren't getting it, then I'm probably not going to either.

I'm just going to move onto the next question for now lol.
 
  • #18
shaools said:
If you guys aren't getting it, then I'm probably not going to either.

I'm just going to move onto the next question for now lol.

Basically the 71 N in the back of the book is wrong, IF you interpret 60 N as being the tension in the rope. If you interpret the 60 N as being the total force acting downward on the upper bucket (including its weight), then the tension in the rope, T, is yet another unknown. This second interpretation means that the 2.2 m/s2 is wrong. They can't both be right.
 
  • #19
Well, I believe that the consensus will be that the "71 N" answer is a "typo" (to be nice to the author). Probably what happened is that the author hastily calculated the answer without themself working the problem out in detail.

I think it is a good idea for you to move on to a different question.
 
  • #20
Dadface said:
For the bottom bucket 60-5*9.8=5a from which a=2.2m/s^2
For the top bucket F-60=5*2.2 from which F=71N
This looks to be how they got the book answer, but it's obviously an error. (The calculation for the top bucket neglects the weight of the bucket.)
 
  • #21
cepheid said:
Then aren't you interpreting the 60 N inconsistently? For the lower bucket, you assume it is the tension in the rope, and doing so gives you a value for a.

For the upper bucket, you suddenly assume that the 60 N is the tension in the rope PLUS the weight of the bucket? No wonder the answer for a is not consistent with the answer for Fapp. See my previous post.

Yes I agree with you and turin.For the whole system F-10*9.8=10*2.2 from which F=120N.
What sort of person jumps in without thinking and ignores the weight of a bucket?Whoops,It was me.
:uhh:
 
Last edited:
  • #22
Ah well I appreciate all of you guys taking the time to look at this one.
It's horrible when the back of the book is wrong :/
 

1. What is Newton's third law of motion?

Newton's third law states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.

2. How does Newton's third law apply to forces between two objects?

When two objects interact, they exert equal and opposite forces on each other. This means that if object A exerts a force on object B, then object B will exert an equal and opposite force on object A.

3. Can you give an example of Newton's third law in action?

One example of Newton's third law is when you push a table. The table exerts an equal and opposite force on you, which is why you feel resistance when pushing against it.

4. How is the force between two objects related to their masses and distance?

According to Newton's law of gravitation, the force between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them. This means that the greater the mass of the objects and the closer they are to each other, the stronger the force between them.

5. Can Newton's third law be applied to non-contact forces?

Yes, Newton's third law can be applied to non-contact forces such as gravitational and electric forces. For example, the Earth exerts a gravitational force on the moon, and the moon exerts an equal and opposite force back on the Earth.

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